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A146968
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Brocard's problem: positive integers n such that n!+1 = m^2.
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11
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OFFSET
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1,1
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COMMENTS
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No other terms below 10^9.
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LINKS
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EXAMPLE
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7! + 1 = 5041 = 71^2, hence 7 is in the sequence. - Klaus Brockhaus, Nov 05 2008
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MATHEMATICA
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Select[Range[10], IntegerQ[Sqrt[#!+1]]&] (* Harvey P. Dale, Jan 31 2015 *)
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PROG
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(Shell) #!/bin/sh n=0 while(true) do n=`echo $n + 1 | bc` calc "($n! + 1)" ^ "(1 / 2)" | grep -v \. done
(Magma) [ <n, p>: n in [1..8047] | t where t, p:=IsSquare(Factorial(n)+1) ]; // Klaus Brockhaus, Nov 05 2008
(PARI) { for (n=1, 60100, if(issquare(n!+1) == 1, print(n) ) ) } \\ Marco Bellaccini (marcomurk(AT)tele2.it), Nov 08 2008
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CROSSREFS
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KEYWORD
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bref,nonn,hard
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AUTHOR
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Marco Bellaccini (marcomurk(AT)tele2.it), Nov 03 2008
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EXTENSIONS
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STATUS
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approved
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